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Magazine Chapter 2: Problem 3
Each of the graphs of the functions has one relative maximum and one relative minimum point. Find these points using the first-derivative test. Use a variation chart as in Example \(1 .\) $$ f(x)=-x^{3}+6 x^{2}-9 x+1 $$
Short Answer
Expert verified
Relative maximum at (3, 1) and relative minimum at (1, -3).
Step by step solution
01
Find the First Derivative
To find the relative maximum and minimum points, first find the derivative of the function. For the function f(x) = -x^3 + 6x^2 - 9x + 1, the derivative is: f'(x) = -3x^2 + 12x - 9. This will help in determining the critical points.
02
Find the Critical Points
Set the first derivative equal to zero to solve for the critical points: -3x^2 + 12x - 9 = 0. Factoring out the quadratic, we get: -3(x^2 - 4x + 3) = 0. Solving for x, we find: (x - 3)(x - 1) = 0. So the critical points are x = 3 and x = 1.
03
Create a Variation Chart
To determine the nature of these critical points, use a variation chart. Evaluate the sign of the first derivative in the intervals determined by the critical points: For x in (-鈭�, 1): Test a point, such as x = 0: f'(0) = -3(0)^2 + 12(0) - 9 = -9 (negative). For x in (1, 3): Test a point, such as x = 2: f'(2) = -3(2)^2 + 12(2) - 9 = 3 (positive). For x in (3, 鈭�): Test a point, such as x = 4: f'(4) = -3(4)^2 + 12(4) - 9 = -9 (negative).
04
Determine Relative Extrema
Using the results from the variation chart, determine the relative extrema. For x = 1: f'(x) changes from negative to positive, thus it is a relative minimum. For x = 3: f'(x) changes from positive to negative, thus it is a relative maximum.
05
Compute Function Values at Critical Points
To find the actual points, compute the function values at the critical points: f(1) = -1^3 + 6(1)^2 - 9(1) + 1 = -1 + 6 - 9 + 1 = -3. f(3) = -3^3 + 6(3)^2 - 9(3) + 1 = -27 + 54 - 27 + 1 = 1.
06
Summarize the Results
The relative maximum point is (3, 1) and the relative minimum point is (1, -3).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points are essential in finding where a function potentially reaches its highest or lowest values locally. To find these, we need to first obtain the first derivative of the function. For the provided function \( f(x) = -x^3 + 6x^2 - 9x + 1 \), the derivative is computed as \( f'(x) = -3x^2 + 12x - 9 \). Once we have the first derivative, we set it equal to zero: \( -3x^2 + 12x - 9 = 0 \). Solving this equation, we factor it into \( (x - 3)(x - 1) = 0 \), giving us the critical points at \( x = 3 \) and \( x = 1 \). Critical points are where the function's slope is zero and thus are candidates for relative maxima or minima.
Relative Maximum
A relative maximum is a point where the function switches from increasing to decreasing. To confirm this, we use the first-derivative test and a variation chart. First, we plug in values into \( f'(x) \) around our critical points. With \( x = 3 \), the derivative changes from positive to negative as we move through the point, showing that \( x = 3 \) is a relative maximum. This means the function attains a peak at this point. We find \( f(3) \) by substituting \( x = 3 \) back into the original function: \( f(3) = -27 + 54 - 27 + 1 = 1 \). The relative maximum point is thus \( (3, 1) \).
Relative Minimum
A relative minimum is a point where the function switches from decreasing to increasing. To determine this, the first-derivative test is again our tool. Next, we analyze \( x = 1 \). Testing values around this point, the derivative changes from negative to positive, confirming \( x = 1 \) is a relative minimum. We now find \( f(1) \) by substituting \( x = 1 \) back into the original function: \( f(1) = -1 + 6 - 9 + 1 = -3 \). So, the relative minimum point is \( (1, -3) \). This is the point where the function reaches a low compared to its neighboring points.
Variation Chart
A variation chart helps visualize how the function behaves across different intervals. Here鈥檚 how to build it:
For example,
By interpreting this chart, you can tell where the function rises or falls, furnishing a clearer picture of its overall behavior and helping identify the relative maxima and minima precisely.
- Identify the intervals formed by the critical points. In this case, they are \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \).
- Test points from each interval in the first derivative \( f'(x) \).
For example,
- For \( x < 1 \), testing \( x = 0 \): \( f'(0) = -9 < 0 \), so the function is decreasing.
- For \( 1 < x < 3 \), testing \( x = 2 \): \( f'(2) = 3 > 0 \), so the function is increasing.
- For \( x > 3 \), testing \( x = 4 \): \( f'(4) = -9 < 0 \), so the function is decreasing.
By interpreting this chart, you can tell where the function rises or falls, furnishing a clearer picture of its overall behavior and helping identify the relative maxima and minima precisely.
